Week 5
Recitation Section
1. The following elements are exceptions to the general rule on orbital filling explain
each
a. Nb41, Pd46
b. Mo42, Ag47
c. Ru44, Rh45
2. Iz: Put in order:
a. Ni,
Ni+
b. Ne, Ar
c. Ne,
Na+
d. F,
Ar
3. Size: Put in order of increasing size
a. N3-, O2-,
F-,
Ne,
2+
3+
b. Mg , Al
c. N,
O,
F
d. F ,
Cl ,
Br+
e. Li , Be
Na+
Lewis Dot Structure Rules:
1. The total number of electrons to use is the sum of the valence electrons for all the
atoms with the charge subtracted from the total. (i.e. SO32- is 6 (for S) + 3*6 (for
O) – (-2) for charge =26 electrons
2. Place the electrons around the atoms so that each atom has 8. Exceptions are H
and Li have 2, B can have 6, and the 3rd row elements can have more than 8.
3. Try to maximize the number of bonds that each atom has.
4. Calculate the partial charges on the atoms (it must add up to the total charge).
5. If there is more then one structure the structure with no partial charges is lowest in
energy.
6. If you must have partial charge put the negative charge on the most
electronegative atom and the positive charge on the least electronegative atoms.
Avoid double negative and double positive atoms if possible.
7. Look for resonance structures.
4. Lewis Dot: Write the Lewis dot structures for:
a. CO, SO2, CO2,
CN-,
SF6 ,
,
b. CH4, H2C2, H4C2, C6H6 , C6H12,
H2SO4,
C8H14
H3PO4
5. Write Resonance and Lewis Dot structures for: NO2-, NO3-, SO2, SO3
6. Oxo Acids:
The four rules of Oxoacids are:
a. As electronegativity of central atom increases the acidity increases
b. As the positive formal charge (no expanded octet) on the central atom
increases the acidity increases
c. As the number of resonance structures of the dissociated acid increase the
acidity increases.
d. As the number of O atoms that do not have an attached H increases acidity
increases.
7. Explain why these rules are true.
8. Put in order of increasing acid strength
a. H3PO4, H2SO4, HClO4
b. write resonance structures for the above conjugate base: H2PO4–, HSO4–,
ClO4–
c. What is the formal charge on the central atom with no expanded octet H3PO3
is a weak acid write a Lewis structure for it and explain why.
9. What are the formal charge on each atom in: SCN-, SCO, N3–
10. Resonance structures: CH3COO–, SCN-,, SO2, SO42–, C6H6,
11. VSEPR:
Define Steric number:
12. Fill in the table below:
Steric #
geometry
Example
6
SF6
5
PF5
4
CH4
3
2
13. Predict the shapes of: CH4, NH3, H2O,
a. why does the angle change
b. H4C2
SO3, SO2, O3
14. Fill in the table:
Steric #
3
3
4
4
4
5
5
5
5
6
6
6
6
6
Geometry of
electrons
Oh
15. Fill in the table:
Lone Pairs
SN
Lone Pairs
0
1
0
1
2
0
1
2
3
0
1
2
3
4
Molecular shape
Molecular Shape
Oh
IF4-,
PF3,
XeF2,
SO2,
SO42-,
PF5,
BF3,
CS2,
SO3,
ICl3,
CBr4,
SiH4,
SF2,
SeF6,
PF3
16. What is the shape around each of the C atoms in H3CCOOH.
17. H3PO3 is a weak acid write a Lewis structure for it and explain why.
Answers:
1. The following elements are exceptions to the general rule on orbital filling explain
each
a. Nb41, Pd46 (steals two 5s electrons to get half filled 4d shell, steals two 5s electrons to get
b.
c.
completely filled 4d shell)
Mo42, Ag47 (the both steal a 5s electron to get half filled or completely filled 4d shell)
Ru44, Rh45 (steal a 5s electron to add another electron to 4d shell)
Remember that the energies of the 4d and 5s are close. As you add electrons and protonss the n-1 shell
become lower in energy faster than the n shell so as you move across the 2 nd transition series the 4d is
dropping in energy faster that the 4s. Thus at some point the 4d orbitals will start to “steal” electrons
from the 5s. Initially it is a comparison of the energy of the 5s orbital plus the energy to pair the
electrons in the s orbital (spin paring energy due to the electron-electron interaction of two electrons in
the 5s) vs the energy of the 4d.
So as we move across at some point it becomes better to put electrons into separate d suborbitals
(no spin pairing energy) then to pair them in the s orbital. As you move further the 4d become even
lower energy and it is better to put an electron in the d rather that the s even if you have to pair then in
the d. So the configurations are as shown
Rb Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
5s1 5s2 5s24d1
5s24d2
5s14d4
5s14d5
5s24d5
5s14d7
5s14d8
4d10 5s14d10
5s24d10
Thus the simple rule that the 5s fills before the 4d has a bunch of exceptions in the second row. You
might want to look at page 551 in OGN to see the exceptions in the 2 nd row.
A semi quantitative explanation (read only if you dare):
Let the energy of the 5s and 4d orbital be E5s and E4d respectively and the energy of spin paring
energy (i.e. The energy to put 2 electrons in the same orbital due to the electron-electron interactions )
be Ps and Pd for the s and d orbitals respectively. Remember that as you move across the Periodic table
(left to right) you find that all the energies of the orbitals decrease However, as we move across the
table (left to right) Ed will decrease faster than Es. We also would expect that P5s < P4d since the 4d is a
smaller orbital than the 5s. The energies for the configurations that we are considering are written
below in the table.
For example consider n=3, Y. If we have a configuration of 5s 24d1 we would have two electrons
in the 5s which gives 2E5s and they are paired so you have Ps and one electron in the 4d so add E4d
which gives 2E5s+Ps+E4d However, if we had a configuration of 5s14d2 we then only have one 5s
electron (E5s) no paired s (or d electrons) and two 4d electrons (2E4d) which gives E5ss+2E4d like wise if
the configuration is a 5s04d3 we have 3 d electrons with no pairing and get 3E4d.
Now the question is which is more stable? The answer is that 5s24d1 is the most stable so that
means that 2E5s+Ps+E4d<E5s+2E4d or E5s+Ps < E4d . Note that this does not say if the 5s or the 4d
orbital is more stable only that the 5s plus spin pair energy is more than the 4d.
If we ask the same question for Zr (n=4) we find that we as have the energy of the 5s orbital plus
the spin pairing energy is less then the energy of the 4d. However the E 4d is going down faster than the
E5s so by n=5 we have that the 5s14d4 is lower that the 5s24d3 configuration. So now we get or E5s+Ps >
E4d
The lower energy configuration is the one shown in bold. As you see there are two factors that are
important the energy of the orbitals and the spin pairing energy and the whole explanation becomes
complicated since the orbital energies are dropping as we move right while the pairing energies are
not.
n=5s+4d
electrons
3, Y
4, Zr
5, Nb
6, Mo
7, Tc
8, Ru
9, Rh
5s24dn-2
5s14dn-1
5s04dn
Reason
2E5s+Ps+E4d
E5s+2E4d
3E4d
E5s+Ps<E4d
2E5s+Ps+3E4d
E5s+4E4d
5E4d
E5s+Ps>E4d
2E5s+Ps+2E4d
E5s+3E4d
4E4d
E5s+Ps<E4d
2E5s+Ps+4E4d
E5s+5E4d
6E4d+Pd
E5s+Ps>E4d
2E5s+Ps+5E4d
E5s+6E4d+Pd
7E4d+2Pd
E5s+Ps<E4d+P4d
2E5s+Ps+6E4d
E5s+7E4d+2Pd
8E4d+3Pd
E5s+Ps>E4d+P4d
2. Iz: Put in order:
d. Ni, Ni+
(Ni+ > Ni Iz(2) always > Iz(1))
e. Ne, Ar
(Ne > Ar move up table the Iz goes up)
+
f. Ne, Na
(Na+ > Ne it has more protons)
g. F, Ar
(F > Ar this is harder to guess but both F and Ar are < Ne
but moving down is more dramatic than moving across the PT).
3. Size: Put in order of increasing size
h. N3-, O2-, F-, Ne, Na+
(-, Na+, Ne, F, O2--, N3-)
i. Mg2+, Al3+
(Al3+ Mg2+)
j. N, O, F
(F, O, N)
k. F , Cl , Br
(F-, Cl-, Br-)
+
l. Li , Be
(Li+, Be)
4. Lewis Dot: Write the Lewis dot structures for:
m. CO, SO2, CO2, CN-, SF6 , H2SO4, H3PO4
:C:::O:,
O–—S+=O  O=S+–O– O=S=O
O=C=O
C–N+
O
OH
O
S2+
O
H
H
O
S
O-
O
O-
O
O
H
O
H
F
F
F
S
F
F
H
O
F
S+
O
H
H
S+
O
O-
O
n. CH4, H2C2, H4C2, C6H6,, C6H12, C8H14
H
H
H
H
C
H
H
C
C
H
H
C
N
C
O
H
H
H
H
H
o.
5. Write Resonance and Lewis Dot structures for: NO2-, NO3-, SO2, SO3
H
H
O
N
O_
-
O
N
O
O
O
-
O
O
O-
O
S
S+
S+
O
O
O
6. Oxo Acids:
The four rules of Oxoacids are:
p. As electronegativity of central atom increases the acidity increases
q. As the positive formal charge (no expanded octet) on the central atom
increases the acidity increases
r. As the number of resonance structures of the dissociated acid increase the
acidity increases.
s. As the number of O atoms that do not have an attached H increases acidity
increases.
7. Explain why these rules are true.
8. Put in order of increasing acid strength
2E5s+Ps+7E4d
E5s+8E4d+3Pd
9E4d+4Pd
E5s+Ps>E4d+P4d
a. H3PO4 < H2SO4 < HClO4
b. write resonance structures for the above conjugate base: H2PO4–, HSO4–,
ClO4–
OH
0
OH
O-
S
-O
S
O
O
P
OH
O
0
OH
-O
O
S2+
O-
O-
O
P
OH
O
O
Çl
OO-
O
Çl
O
Cl+
O
-O
O
O
O-
0
0
O
O-
O
Çl
0
O
0
O-
0
Cl+
0
Çl
O
0
O0
S
S-
OH
-O
O-
0
OH
O-
O
OH
0
0
OH
Cl-
O
O
O-
Cl+
O-
-O
Cl+
O
.
c. What is the formal charge on the central atom with no expanded octet (0,
2+, 3+)
9. Formal charge: SCN-, SCO, N3– S=C=N–, S=C=O,
N–=N2+=N–N2–-N+NNN+-N2–
10. Resonance structures: CH3COO–, SCN-,, SO2, SO42–, C6H6,
O-
O
H
C
H
OH
H
0
0
O
C
O
O-
S
OS
O-
VSEPR:
O
S
C
+O
NS-
O
S-
O
OO
0
SO
O+
+O
S2O
OO
-O
S2+
O-
O-
0
S2O
O+
O
11. Define Steric number:
12. Fill in the table below:
Steric #
6
5
4
3
2
geometry
Oh
Trigonal bipyramidal
Td
triangle
linear
example
SF6
PF5
CH4
13. Predict the shapes of:
CH4, NH3, H2O,
why does the angle change
H
109.5
C
H
O
H
H
H
H
N
H
107
H
H
105
angle decreases
because the lone pairs take up more space and push bonding pairs away.
H4C2 SO3, SO2, O3
120
H
H
O
C
S
120
120
C
S
H
H
O
O
O
O
<120
O
O
<120 O
14. Fill in the table:
Steric #
3
3
4
4
4
5
5
5
5
6
6
6
6
6
Geometry of electrons
Trigonal planar
Trigonal planar
Td
Td
Td
Trigonal bipyramidal
Trigonal bipyramidal
Trigonal bipyramidal
Trigonal bipyramidal
Oh
Oh
Oh
Oh
Oh
15. Filll in the table
Lone Pairs
SN
2
IF4 ,
6
Lone Pairs
0
1
0
1
2
0
1
2
3
0
1
2
3
4
Molecular shape
sq planar
Molecular Shape
Trigonal planar
linear
Td
Trigonal pyramidal
bent
Trigonal bipyramidal
sawhorse
T shape
linear
Oh
Sq pyramided
Sq planar
Bent T
linear
PF3,
XeF2,
SO2,
SO42-,
PF5,
BF3,
CS2,
SO3,
ICl3,
CBr4,
SiH4,
SF2,
SeF6,
PF3
4
5
3
4
5
3
2
3
5
4
4
4
6
4
1
3
1
0
0
0
0
0
2
0
0
2
0
1
Trigonal pyramid
linear
bent (angle <120)
Td
trigonal bipyramidal
trigonal planar
linear
trigonal planar
trigonal planar
Td
Td
bent
Oh
trigonal pyramidal
16. What is the shape around each of the C atoms in H3CCOOH.
Td, trigonal planar
17. H3PO3 is a weak acid write a Lewis structure for it and explain why.